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• Arsalan Asif

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MS-291: Engineering Economy

(3 Credit Hours)

What is Next!!!! Week Step I

Step 2

Contents

Reading Materials

Foundation of Engineering Economy, various cost concepts

Chapter 1: (B&T 2012)

Factors: How time and Interest Affect Money

Chapter 2: (B&T 2012)

Nominal and Effective Interest Rates

Chapter 3: (B&T 2012)

Combining Factors and Spreadsheet Functions

Chapter 4: (B&T 2012)

Present Worth Analysis

Chapter 5 : (B&T 2012)

Mid-Term Step 2

Step 3

Reading Materials

Annual Worth Analysis

Chapter 6: (B&T 2012)

Rate of Return Analysis: Single/ Multiple Alternatives

Chapter 7 & 8: (B&T 2012)

Benefit/Cost Analysis and Public Sector Economics

Chapter 9: (B&T 2012)

Independent Projects with Budget Limitations

Chapter 12: (B&T 2012)

Project Financing and Noneconomic Attributes

Chapter 10: (B&T 2012)

Breakeven and Payback Analysis

Chapter 13: (B&T 2012)

Effects of Inflation

Chapter 14: (B&T 2012)

Depreciation Measures

Chapter 16: (B&T 2012)

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Summary so for… Present Worth Convert all cash flows to a single sum equivalent at time zero using the MARR Future Worth Convert all cash flows to a single sum equivalent at the end of the planning horizon using the MARR Capitalized Worth Method Determine the single sum at time zero that is equivalent at i=MARR to a cash flow pattern that continues indefinitely

Chapter 6 Annual Worth Analysis Engineering Economy

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Content of this Chapter 1)

Annual Worth Basics

2)

Advantages and Uses of AW

3)

Capital Recovery and AW values

4)

AW analysis

5)

Perpetual life

6)

Life-Cycle Cost analysis

Annual Worth(AW): Basics • Annual worth is an equal periodic series of dollar amounts that is equivalent to the cash flows (inflows & outflows), at MARR (or interest rate) • It can be interchangeably use for Annuity “A” we studied earlier • AW, PW and FW has the following relationship: AW = PW (A/P, i, n) = FW(A/F, i, n) • The n in the factors is the number of years for equal-service comparison. This is the LCM or the stated study period of the PW or FW analysis.

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Example: Different-Life Alternatives PW (Example 5.3 used in Example 6.1) National Homebuilders, Inc., plans to purchase new cut-and-finish equipment. Two manufacturers offered the estimates below. First cost, $ Annual cost, $/year Salvage value, $ Life, years (a)

Vendor A 15,000 3,500 1,000 6

Vendor B 18,000 3,100 2,000 9

Determine which vendor should be selected on the basis of a present worth comparison, if the MARR is 15% per year.

(b) National Homebuilders has a standard practice of evaluating all options over a 5year period. If a study period of 5 years is used and the salvage values are not expected to change, which vendor should be selected? We did this example in Last chapter.

Example 6.1 National Homebuilders, Inc. evaluated cut-and-finish equipment from vendor A (6-year life) and vendor B (9-year life) (In Example 5.3). The PW analysis used the LCM of 18 years. Consider only the vendor A option now. Consider the cash flows for all three life cycles, 18 years… (first cost $15,000; annual M&O costs $3500; salvage value $1000). Demonstrate the equivalence of PW over three life cycles and AW over one cycle at i 15% . In Example 5.3, present worth for vendor A was calculated as PW $45,036.

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Example 6.1 PW = $ 45036

Awfirst cycle = ― 15000(A/P, i, n)+1000(A/F, i, n) ― 35000

i=15%

AW = ― 15000(A/P, 15%, 6)+1000(A/F, 15%, 6) ― 35000 AW = ― $7349

3 life cycles

0 1 2 3 4 5 6 7

8 9 10 11 12 13 14 15 16 17 18

$1,000 0

1

2

3

4

5

But we know that PW of alternative A for 18 years is given as: ― $45,036

6 Life cycle 1

$1,000 $3500 0

1

2

3

4

5

AW = ― 45,036(A/P, i, n)

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Life cycle 2

$15,000

$1,000

$3500

$15,000

0

1

2

3

4

5

6

Life cycle 3

AW = ― 45,036(A/P, 15%, 18) AW = ― $7349

$3500

$15,000 0 1 2 3 4 5 6 7

8 9 10 11 12 13 14 15 16 17 18 19 20….

The one-life-cycle AW value and the AW value based on 18 years (3 life-cycle) are equal.

AW = ‒$ 7349

Important!!!!! • The AW value determined over one life cycle is the AW for all future life cycles. • Therefore, it is not necessary to use the LCM of lives to satisfy the equal-service requirement • But its important to remember, that AW method is also based on assumption of equal service requirement or equal life assumption.

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Annual Worth(AW): Basics • Other names for AW are equivalent annual worth (EAW), equivalent annual cost (EAC), annual equivalent (AE), and equivalent uniform annual cost (EUAC) • AW and PW (and all other alternative evaluation methods such as, FW, BCR, etc also) select same alternatives , provided they are performed correctly

AW Method: Assumption 1. 2. 3.

The services provided are needed for at least the LCM of the lives of the alternatives. The selected alternative will be repeated for succeeding life cycles in exactly the same manner as for the first life cycle. All cash flows will have the same estimated values in every life cycle. Exactly the same as for PW method!!!!

• If first two assumptions are not reasonable, a study period must be used • for third assumption, if cash flows are expected to change exactly with the inflation (or deflation) rate its Okay….if not…., new cash flow estimates must be made for each life cycle, and again a study period must be used

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Advantages and Uses of Annual Worth • AW method offers a prime computational and interpretation advantage because the AW value needs to be calculated for only one life cycle • AW is not necessary to use the LCM of lives to satisfy the equalservice requirement because AW for one life cycle is the AW for all future life cycles • AW more desirable compared to Present worth because: i. AW is easy to calculate; ii. AW is easy to understand (express in $ per year) iii. Its assumptions are identical to those of the PW method(such as equal life services or cash flow same in each life cycle)

Calculation of Annual Worth (AW) An alternative should have the following cash flow estimates: • Initial investment P. This is the total first cost of all assets and

services required to initiate the alternative (some time this investment take place overall several years)

• Salvage value S. This is the terminal estimated value of assets at the

end of their useful life. • The S is zero if no salvage is anticipated; S is negative when it will cost money to dispose of the assets. • For study periods shorter than the useful life, S is the estimated market value or trade-in value at the end of the study period.

• Annual amount A. This is the equivalent annual amount (costs only for cost alternatives; costs and receipts for revenue alternatives). Often this is the annual operating cost (AOC) or M&O cost, so the estimate is already an equivalent A value.

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Calculation of Annual Worth (AW) AW = CR + A Annual Worth is composed of two components 1. Capital Recovery for initial investment (at MARR) 2. Equivalent Annual Amount, A

Capital Recovery Amount • Capital recovery (CR) is the equivalent annual amount that the asset, process, or system must earn (new revenue) each year to just recover the initial investment plus a stated rate of return over its expected life • Any expected salvage value is considered in the computation of CR CR = ― P (A/P,i,n ) + S (A/F,i,n ) or CR = ―(P ―S)[A/P, i, n) ― S(i)]

There is certain derivation how to get the second formula…but we do not need that.

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Capital Recovery Amount CR = ― (P ―S)[A/P, i, n) + S(i)] There is a basic logic to this formula • Subtracting S from the initial investment P before applying the A/P factor recognizes that the salvage value will be recovered, this reduces CR, the annual cost of asset ownership •

But S is not recovered until year n of ownership is compensated for by charging the annual interest S ( i ) against the CR

Example: AW, CR Lockheed Martin is increasing its booster thrust power in order to win more satellite launch contracts from European companies interested in opening up new global communications markets. A piece of earth-based tracking equipment is expected to require an investment of $13 million, with $8 million committed now and the remaining $5 million expended at the end of year 1 of the project. Annual operating costs for the system are expected to start the first year and continue at $0.9 million per year. The useful life of the tracker is 8 years with a salvage value of $0.5 million. Calculate the CR and AW values for the system, if the corporate MARR is 12% per year.

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Example: AW, CR Capital Recovery in millions is given as

CR = ― P (A/P,i,n ) + S (A/F,i,n ) P = 8 + 5(P/F, i, n) = 8+5(P/F, 12%, 1) = $12.46

Now CR = ― P(A/P, i, n) + S(A/F, i, n) CR = ― 12.46(A/P, 12%, 8) + 0.5(A/F, 12%, 8) CR = ― 12.46(0.20130) + 0.5(0.08130) CR = ― $2.47

Interpretation: It means that each and every year for 8 years, the equivalent total net revenue from the tracker must be at least $2,470,000 just to recover the initial present worth investment plus the required return of 12% per year. This does not include the AOC of $0.9 million each year.

Example: AW, CR Annual worth: To determine AW, the cash flows must be converted to an equivalent AW series over 8 years

AW = CR + A Now, AW = ? CR = -$2.47 and A = -$0.9

AW = -2.47 + (- 0.9) AW = ― $3.37 million per year

Interpretation : $3.37 million is the AW for all future life cycles of 8 years, provided the costs rise at the same rate as inflation, and the same costs and services are expected to apply for each succeeding life cycle.

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Annual Worth Analysis: Mutually Exclusive Events • Convert all cash flows to equivalent uniform annual amounts over the planning horizon using the MARR For mutually exclusive alternatives, whether cost- or revenuebased, the guidelines are as follows: • One alternative: If AW ≥ 0, the alternative is economically justified • Two or more alternatives: Select the alternative with the AW that is numerically largest, that is, less negative or more positive. • This indicates a lower AW of cost for cost alternatives or a larger AW of net cash flows for revenue alternatives.

Annual Worth Analysis: Independent Events • If the projects are independent, the AW at the MARR is calculated. • All projects with AW ≥ 0 are acceptable.

Not necessary to use LCM for different life alternatives while using AW (does not matter if its mutually exclusive events or independent events)

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Class Practice: Comparing Alternatives Using AW criteria TT Racing and Performance Motor Corporation wishes to evaluate two alternative CNC machines for NHRA engine building. Use the AW method at 10% per year to select the better alternative. First Cost, $ Annual operating cost, $/year Salvage value, $ Life, years

10%

Machine R -250,000 - 40,000 20,000 3

Single Payments

Machine S -370,000 -50,000 30,000 5

Uniform Series Factors

n

Compound Amount (F/P)

Present Worth (P/F)

Sinking Fund (A/F)

Capital Recovery (A/P)

3

1.3310

0.7513

0.30211

0.40211

5

1.6105

0.6209

0.16380

0.26380

15

4.1772

0.2394

0.03147

0.13147

Class Practice: Comparing Alternatives Using AW criteria TT Racing and Performance Motor Corporation wishes to evaluate two alternative CNC machines for NHRA engine building. Use the AW method at 10% per year to select the better alternative. First Cost, $ Annual operating cost, $/year Salvage value, $ Life, years

Machine R -250,000 - 40,000 20,000 3

Machine S -370,000 -50,000 30,000 5

Solution: What if we want to compare on PW criteria ? AWR = – 250,000(A/P,10%,3) – 40,000 + 20,000(A/F,10%,3) = – 250,000(0.40211) – 40,000 + 20,000(0.30211) = $ – 134,485 AWS = – 370,500(A/P,10%,5) – 50,000 + 30,000(A/F,10%,5) = – 370,500(0.26380) – 50,000 + 30,000(0.16380) = $ – 142,824

Select alternative R

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Annual Worth Analysis for Permanent Investment • We may have a permanent Investment (such as public sector projects) for which we need evaluation based on Annual Worth • In such case; convert cash flows (recurring & nonrecurring) to equivalent uniform annual amounts A for one cycle • This automatically annualizes them for each succeeding life cycle. • Add all the A values to the CR amount to find total AW (as AW = CR + A, where CR = ― P (A/P,i,n ) + S (A/F,i,n ))

Question 6.26: Annual Worth Analysis for Permanent Investment Compare two alternatives for a security system surrounding a power distribution substation using annual worth analysis and an interest rate of 10% per year. First Cost, $ Annual operating cost, $/year Salvage value, $ Life, years

C ― 25,000 ― 9,000 3,000 3

AWC = ―25,000(A/P,10%,3) – 9000 + 3000(A/F,10%,3) = ― 25,000(0.40211) – 9000 + 3000(0.30211) = $ ― 18,146 per year AWT = ― 130,000(0.10) – 2500 = $ ― 15,500 per year Select the T system

T ― 130,000 ― 2,500 150,000 ∞

How to find present worth for undefined life alterative ?

( CC = A / i) ( AW = CC x i)

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Life Cycle Cost Analysis Life Cycle Cost …Sound Familiar ?

Life-cycle Costs (Old Slide) • Life-cycle - all the time from the initial conception of an idea to the death of a product (process) • Life-cycle costs - sum total of all the costs incurred during the life cycle • Life-cycle costing - designing a product with an understanding of all the costs associated with a product during it’s life-cycle

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Product Life-cycle (Old Slide) Begin Needs assessment and justification

Time

Conceptual or preliminary design phase

Impact Analysis Requirements Overall Feasibility Conceptual Design Planning

Proof of concept Prototype Development and testing Detailed design planning

Detailed design phase

Allocation of resources Detailed specification Component and supplier selection Production or construction phase

Production or Construction Phase Product, goods and service built All supporting facilities built Operation al use planning

End

Operational Phase

Decline and retirement phase

Operational Use Use by ultimate customer Maintenance and support Process, materials and methods use Declined and retirement planning

Decaling Use Phase out Retirement Responsible disposal 1-29

Life-Cycle Cost Analysis • Life-cycle cost (LCC) analysis use Annual Worth or Present Worth methods to evaluate cost estimates for the entire life cycle of one or more projects • Estimates will cover the entire life span from the early conceptual stage, through the design & development stages, throughout the operating stage, and even the phase out and disposal stages. • Both direct and indirect costs are included to the extent possible, and differences in revenue and savings projections between alternatives are included.

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Life-Cycle Cost Analysis • Direct costs refers to material, human labor, equipment, supplies, and other costs directly related to a product, process, or system • Indirect cost refers to taxes, management, legal, warranty, quality, human resources, insurance, software, purchasing, etc.

Life-Cycle Cost Analysis • Life-cycle cost (LCC) analysis use Annual Worth or Present Worth methods to evaluate cost estimates for the entire life cycle of one or more projects • Most commonly the LCC analysis includes costs, and the AW method is used for the analysis, especially if only one alternative is evaluated • If there are expected revenue or other benefit differences between alternatives, a PW analysis is recommended. • Public sector projects are usually evaluated using a benefit/cost analysis (Chapter 9), rather than LCC analysis, because estimates to the citizenry are difficult to make with much accuracy

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Life-Cycle Cost Applications • Life-Cycle Costs applications are: life-span analysis for military & commercial aircraft, new manufacturing plants, new automobile models, new and expanded product lines, and government systems at federal and state levels • For example, the U.S. Department of Defense requires that a government contractor include an LCC budget and analysis in the originating proposal for most defense systems

When to use Life-Cycle Costs ? • Life-Cycle Costs analysis is more effective when significant percentage of the life span (post purchase) costs, (relative to the initial investment) is expected to be spend on operating, maintenance, and similar costs once the system is operational. • Example 1: The evaluation of two equipment purchase alternatives with expected useful lives of 5 years and M&O costs of 5% to 10% of the initial investment does not require an LCC analysis. •

Example 2: Exxon-Mobil wants to evaluate the design, construction, operation, and support of a new type and style of tanker that can transport oil over long distances of ocean. If the initial costs are in the $100 millions with support and operating costs ranging from 25% to 35% of this amount over a 25-year life, the logic of an LCC analysis will offer a better understanding of the economic viability of the project.

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How Life-Cycle Cost Analysis Works • To do LCC analysis one need to identify each stage of product, system or an alternative during entire life • However, at this stage, we will be given with all stages of life cycle & we will asked to calculate LCC. We, then follow the following steps: 1. Calculate the PW by phase and stage 2. Add all PW values of each stage 3. Using PW calculated in Step 2,…Calculate the AW over entire life-cycle of the alternative

Practice Question: Life-Cycle Cost Analysis A medium-size municipality plans to develop a software system to assist in project selection during the next 10 years. A life-cycle cost approach has been used to categorize costs into development, programming, operating, and support costs for each alternative. There are three alternatives under consideration, identified as M, N, and O. The costs are summarized below. Use an annual life-cycle cost approach to identify the best alternative at 8% per year.

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8% Life-Cycle Uniform Series Practice Question: CostFactors n Capital Present Worth Analysis Recovery (A/P) (P/A)

2 3 4 5 10

0.56077 0.38803 0.30192 0.25046 0.14903

1.7833 2.5771 3.3121 3.9927 6.7101 Single Payments

8% n

Compound Amount (F/P)

Present Worth (P/F)

2

1.1664

0.8573

3

1.2597

0.7938

4

1.3605

0.7350

5

1.4693

0.6806

10

2.1589

0.4632

Question 6.33: Life-Cycle Cost Analysis PWM = – 250,000 – 150,000(P/A,8%,4) – 45,000 – 35,000(P/A,8%,2) – 50,000(P/A,8%,10) – 30,000(P/A,8%,5) PWM = – 250,000 – 150,000(3.3121) – 45,000 – 35,000(1.7833)-50,000(6.7101) – 30,000(3.9927) PWM = $ – 1,309,517 Annual LCCM = – 1,309,517(A/P,8%,10) = – 1,309,517(0.14903) = $-195,157 PWN = -10,000 – 45,000 - 30,000(P/A,8%,3) –80,000(P/A,8%,10) – 40,000(P/A,8%,10) PWN = – 10,000 – 45,000 - 30,000(2.5771) – 80,000(6.7101) - 40,000(6.7101) PWN = $ – 937,525 Annual LCCN = – 937,525(A/P,8%,10) Annual LCCN = – 937,525(0.14903)

Annual LCCN = $ – 139,719

Select Alternative N

Annual LCCO = $ – 175,000

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THANK YOU

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